I was doing my homework on Polynomials. Something strike my mind about what will be the solution for a polynomial with degree 4. I just took an example sum, which is below:
f(x) = x⁴ + 2x³ + x² - 2x + 1
I tried many methods but couldn't find the solution. Please help me solving it. Also tell whether there is any relationship between their zeroes and coefficients.
This is not an answer. This is a long-winded comment.
You are given $~f(x) = x^4 + 2x^3 + x^2 - 2x + 1,~$ and you wish to solve the equation $~f(x) = 0.~$ As has already been mentioned, one approach is the rational root theorem.
Because of the unusual nature of this function, there is another approach.
In the polynomial $~f(x),~$ let $~a_k~$ denote the coefficient of $~x^k~: ~k \in \{0,1,2,3,4\}.~$ Then, you have that $~a_4 = a_0,~$ and $~a_3 = \pm a_1.~$ I refer to such a quartic polynomial as a symmetric quartic. There is a special strategy that can be applied to any symmetric quartic.
Construct $~g(x) = \dfrac{f(x)}{x^2} = x^2 + 2x + 1 - \dfrac{2}{x} + \dfrac{1}{x^2}.~$ Note that $~x = 0~$ can not be a solution to the equation $~f(x) = 0.~$ Therefore,
$$f(x) = 0 \iff g(x) = 0.$$
Given any symmetric quartic equation $~f(x) = 0,~$ the equation can be attacked by instead attacking the constructed equation $~g(x) = 0.~$
Depending on whether $~a_3 = a_1,~$ or $~a_3 = -a_1,~$ you would attack $~g(x) = 0,~$ by one of the two substitutions:
$w = \left( ~x + \dfrac{1}{x} ~\right)$
or $~w = \left( ~x - \dfrac{1}{x} ~\right).~$
In this case, you want the latter substitution.
Note that here,
$$w^2 = x^2 - 2 + \frac{1}{x^2} \implies x^2 + \frac{1}{x^2} = w^2 + 2.$$
Then,
$$g(x) = w^2 + 2w + 3.$$
Thus, you have converted the polynomial $~g(x)~$ into a 2nd degree polynomial in the variable $~w,~$ and you can then apply the formula for a quadratic equation against the 2nd degree polynomial equation, $~w^2 + 2w + 3 = 0.$
Once this is done, you will (potentially) obtain two roots of this 2nd degree polynomial equation, $~q_1,~q_2.$
Then, to obtain the roots of the equation $~f(x) = 0,~$ you would solve (separately), the two distinct quadratic equations,
$\displaystyle w = q_1 = x - \frac{1}{x} \implies x^2 - q_1x - 1 = 0$
and $~\displaystyle w = q_2 = x - \frac{1}{x} \implies x^2 - q_2x - 1 = 0.$
According to rational root theorem, every rational root of a polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_0$, where $\forall i\ a_i \in \mathbb Z$ and $a_n \ne 0$ and $a_n \ne 0$ looks like $\dfrac{p}{q}$, where $p$ and $q$ are coprimes (i.e. $\gcd(p, q) = 1)$, $p|a_0$, $q|a_n$. In your case, $a_0 = a_n = 1$, meaning the only possible rational roots are $1$ and $-1$. It is clear, that none of them are roots, meaning there are no rational roots for this polynomial. This usually means, that the roots are really difficult to guess, so to find them you will need some pretty sophisticated methods like Ferrari's method (you can read this wikipedia page for some ideas).
